First slide

Wave nature of matter

Question

An electron of mass m_eand a proton of mass m_p are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an
electron to that associated with proton is

Easy

A

1
B

$m_{p} / m_{e}$
C

$m_{e} / m_{p}$
D

$\sqrt{m_{p} / m_{e}}$

Solution

If q is the charge on the particle and V the potential difference through which it is accelerated, then
$q V = \frac{1}{2} m v^{2} or m v = \sqrt{2 m q V}$
de Broglie's wavelength,
$λ = \frac{h}{m v} = \frac{h}{\sqrt{2 m q V}} ∴ \frac{λ_{e}}{λ_{p}} = \sqrt{\frac{m_{p}}{m_{e}}}$

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